Photon entanglement: why three angles?

[Jabbu]

If you found 100% correlation in a coin flip, then you'd be surprised and wonder what causes the coins to always show heads or tails together. But that's not necessarily true of particles emitted from a common source, since the correlation can result from the event itself, e.g., conservation of angular momentum.

It looks like only two of us are talking about the same thing. But hold on a second. Even if entangled photons have the same polarization relative to their polarizers, only their probability is the same, but it's the polarizers that will ultimately decide what outcome will actually be, and according to probability theory and local causality that's two separate independent 50% probabilities, so there should not be more than 50% matching pairs in the long run.

Therefore, there is no local or classical explanation for 100% correlation. If there was, QM explanation would be superfluous. Any correlation less than 100% can only be less convincing as it is closer to classical prediction.

 

[Jabbu]

I gave you one.

Anyway, your original question was: why do we need to consider more than one angle? The answer is: Because otherwise, there is no known proof that the results cannot be obtained by a local realistic theory.

If you think your locally causal magic trick explanation is really plausible, then it can just the same disprove any other non-local correlation for any other angle or combination of angles. When experiments are performed with multiple angles the data is recorded separately for each one and the result has to be the same as if each angle was tested independently by itself.

 

[DrChinese]

It looks like only two of us are talking about the same thing. But hold on a second. Even if entangled photons have the same polarization relative to their polarizers, only their probability is the same, but it's the polarizers that will ultimately decide what outcome will actually be, and according to probability theory and local causality that's two separate independent 50% probabilities, so there should not be more than 50% matching pairs in the long run.

The above statements are both ambiguous and relating to different subjects. No one can really be sure what you are discussing at any particular point in time. Are you discussing entangled pairs? Are you talking about what actually happens?

Because when you say "probability theory and local causality that's two separate independent 50% probabilities" you are talking about things that are not applicable. Entangled particles are part of a SINGLE system. They do not follow "product state" (separable) statistics.

 

[stevendaryl]

If you think your locally causal magic trick explanation is really plausible

No, I don't think it's plausible, but it is possible. It is enough to show that there is no proof in the case of a single angle.

Look, this is not complicated: There is a proof in the case of three angles. There is no proof in the case of a single angle. That's all there is to it.

 

[stevendaryl]

If you think your locally causal magic trick explanation is really plausible, then it can just the same disprove any other non-local correlation for any other angle or combination of angles.

That's just not true. It works for one angle, but not for a combination of angles.

Let's take the photon EPR case with three angles:
\theta=0, \theta = 60, \theta = 120

A local deterministic hidden variables model would have a parameter \lambda that would take on 8 different values:

\lambda_{+++}
\lambda_{++-}
\lambda_{+-+}
\lambda_{-++}
\lambda_{+--}
\lambda_{-+-}
\lambda_{--+}
\lambda_{---}

If the hidden variable \lambda had value \lambda_{+++}, then the photon will pass through a filter oriented at \theta = 0, \theta = 60, \theta =. If it had value \lambda_{++-}, then the photon will pas through a filter oriented at \theta = 0, \theta = 60 but would be blocked by a filter at \theta = 120. Etc.

We can prove that there does not exist 8 probabilities
P_{+++}, P_{++-}, ... (where P_{+++} is the probability that \lambda has value \lambda_{+++}, etc.) such that this sort of local realistic theory reproduces the predictions of QM.

Now, if we only have a single angle, \theta=0, then the predictions of QM are that
50% of the time, the photon passes, and 50% of the time, it is blocked. It's EASY to come up with a local realistic model for this case. In this case, there are two possible values for \lambda:

\lambda_{+}
\lambda_{-}

The probabilities for these two values are:

P_{+} = 50%, P_{-} = 50%

That trivial model reproduces the QM predictions for a single angle.

 

[Jabbu]

If the hidden variable \lambda had value \lambda_{+++}, then the photon will pass through a filter oriented at \theta = 0, \theta = 60, \theta =. If it had value \lambda_{++-}, then the photon will pas through a filter oriented at \theta = 0, \theta = 60 but would be blocked by a filter at \theta = 120. Etc.

We can prove that there does not exist 8 probabilities
P_{+++}, P_{++-}, ... (where P_{+++} is the probability that \lambda has value \lambda_{+++}, etc.) such that this sort of local realistic theory reproduces the predictions of QM.

Correlation is not calculated between readings for different angles. Experiments with three angles produce three separate data streams, one for each angle, and correlation is calculated independently for each data stream of AB pairs. For theta= 0, Alice and Bob readings should be 100% correlated just as if theta= 0 was tested independently by itself. Other angles have their completely separate data streams of AB pairs with their own probabilities and their own correlations according to QM's cos^2(theta).

Now, if we only have a single angle, \theta=0, then the predictions of QM are that
50% of the time, the photon passes, and 50% of the time, it is blocked. It's EASY to come up with a local realistic model for this case. In this case, there are two possible values for \lambda:

Proposed local classical equation must be confirmed by classical experiments, for any angle, it has to be able to actually work for photon-polarizer interaction in general, or it simply is not true. It has to give the same result as Malus law, it thus can not be anything else. Either Malus law works or the explanation is non-local, there is no other choice compatible with all the other classical experiments.

 

[RUTA]

Correlation is not calculated between readings for different angles. Experiments with three angles produce three separate data streams, one for each angle, and correlation is calculated independently for each data stream of AB pairs. For theta= 0, Alice and Bob readings should be 100% correlated just as if theta= 0 was tested independently by itself. Other angles have their completely separate data streams of AB pairs with their own probabilities and their own correlations according to QM's cos^2(theta).

"theta" is the difference between the two polarizer/SG magnet settings, so whatever the settings of the two detectors, if they are the same, theta = 0. Read the attached paper and go through all the calculations. There are some mistakes, but they're easy to find. After completing this homework assignment, you'll understand how an experiment showing the violation of Bell's inequality is actually done, to include data analysis. After you've finished, come back with your questions.

 

[stevendaryl]

Proposed local classical equation must be confirmed by classical experiments, for any angle, it has to be able to actually work for photon-polarizer interaction in general, or it simply is not true. It has to give the same result as Malus law, it thus can not be anything else. Either Malus law works or the explanation is non-local, there is no other choice compatible with all the other classical experiments.

I just showed you that a local hidden variables theory can reproduce exactly the predictions of QM for a single filter setting (both Alice and Bob have the filters at the same angle). So what you're saying is false: the predictions for a single angle does not demonstrate that anything nonlocal is going on. In contrast, when you have 3 filter settings to choose from, it is not possible to reproduce the predictions of QM by any local theory.

I don't know why you keep bringing up Malus' law. It is completely useless in predicting correlations for the case of entangled photons, because you can't apply it unless you know the photon polarizations.

 

[Jabbu]

"theta" is the difference between the two polarizer/SG magnet settings, so whatever the settings of the two detectors, if they are the same, theta = 0. Read the attached paper and go through all the calculations. There are some mistakes, but they're easy to find. After completing this homework assignment, you'll understand how an experiment showing the violation of Bell's inequality is actually done, to include data analysis. After you've finished, come back with your questions.

I see. You are not surprised correlation for theta= 0 is 100%, as if that is normal. Instead, what you find more convincing the interaction is non-local is that some combined thetas correlation function "S" is greater than 2. Is that it?

 

[RUTA]

I see. You are not surprised correlation for theta= 0 is 100%, as if that is normal. Instead, what you find more convincing the interaction is non-local is that some combined thetas correlation function "S" is greater than 2. Is that it?

Did you read the paper and reproduce all the calculations? I can help you understand the equations and how they are related to the experiment. What you choose to believe about those equations and the experiment is up to you.

 

[morrobay]

The first column of this example above should be read as "for the first pair if we measure the polarization on angle A the left-hand photon will pass and the right-hand one will not; if we measure on angles B or C the left-hand photon will not pass and the right-hand one will". The second column should be read as "for the second pair if we measure the polarization on angle B the left-hand photon will not pass and the right-hand one will; if we measure on angles A or C the left-hand one will pass and the right-hand one will not". This is exactly the local realistic theory I describe above - both photons are created with definite polarization values at all three angles.

You calculate the results by choosing any two of the three possible results because we only get to make two measurements, one on each photon. The challenge is to construct a data set that will lead to a violation of Bell's equality no matter which measurements we choose to make on each pair - and if you try it you'll find that it cannot be done. Therefore, no theory in which the results of measurements on all three angles are predetermined can match the experimental results.

The data set for above is for Alice:
a: ++-+--+
b: --+-++-
c: -+-++-+

I have found a violation from data set from above post. Converting above data to horizontal format including Bobs values:

..A....B
a b c...a b c
+ - -...- + +
+ -+...- + -
- + -...+ - +
+ - +...- + -
- + +...+ - -
- + -...+ - +
+ - +...- + -

n[a+c+] + n[b+c-] ≥ n[a+b+]

n=1 + n=1 ≥ n= 4 violation

This form of inequality is taken from http://math.ucr.edu/home/baez/physics/Quantum/bells_inequality.html

 

[Jabbu]

Did you read the paper and reproduce all the calculations? I can help you understand the equations and how they are related to the experiment. What you choose to believe about those equations and the experiment is up to you.

I did read it. I didn't reproduce calculations, but I promise I will. You neither confirmed nor denied what I said in my last two post. Please be more specific and at least tell me what are we talking about now.

2 < S = E(a,b) - E(a,b') + E(a',b) + E(a',b')

Is that it, the proof experiment data describe non-local interaction?

 

[RUTA]

I did read it. I didn't reproduce calculations, but I promise I will. You neither confirmed nor denied what I said in my last two post. Please be more specific and at least tell me what are we talking about now.

2 < S = E(a,b) - (a,b') + (a',b) + (a',b')

Is that it, the proof experiment data describe non-local interaction?

If all you're trying to understand are the "non-local" implications of entanglement, then any of Mermin's papers will do. Essentially, it comes down to "no instruction sets." The 1994 paper I posted earlier is the easiest to understand in my opinion because it doesn't involve statistics or inequalities.

If you're trying to understand how data is actually taken concerning "non-locality," then read and work through all the calculations in the last paper I posted. In that case, "non-locality" is represented by violating |S| <= 2. I posted that paper because you seemed confused as to what the cosine squared of theta actually means and it's explained nicely in that article.

In either case, all the material you need to understand what physicists mean by "quantum non-locality" has been posted and explained in this thread. If there is something about the equations you don't understand, people here will help you. If you want to argue about semantics or what physicists should or should not believe concerning ontology, the moderators will likely close the thread.

 

[DrChinese]

Is that it, the proof experiment data describe non-local interaction?

No. It is proof that it is non-local or non-realistic. Or both. No one knows for sure.

 

[RUTA]

No. It is proof that it is non-local or non-realistic. Or both. No one knows for sure.

Careful here, Jabbu. This is a more precise use of the term "non-local" as regards entanglement. The term "quantum non-locality" has come to mean "measurement outcomes with entangled particles that violate classical intuition." One way to characterize exactly what it is in classical intuition that is violated by these QM measurement outcomes is to say they are "non-local" and/or "non-realistic." In this sense, "non-local" means they involve "superluminal influences." In the first 1:30 of this talk by Weinstein , for example, he uses "outcome independence" and "parameter independence" instead of "non-local" and "non-real." He then says "the conjunction of these two is sometimes referred to as strong locality or just locality."

 

[Jabbu]

If you're trying to understand how data is actually taken concerning "non-locality," then read and work through all the calculations in the last paper I posted. In that case, "non-locality" is represented by violating |S| <= 2. I posted that paper because you seemed confused as to what the cosine squared of theta actually means and it's explained nicely in that article.

That. So I'm looking at page 907 and I'm trying to calculate their example:

(Ex) a= -45, a'= 0, b= -25.5, b'= 25.5

...they say result is obtained with equations 10, 20, and 21:

(10) Pvv(a,b) = cos^2(b-a) / 2

(20) E(a,b) = Pvv(a,b) + Phh(a,b) - Pvh(a,b) - Phv(a,b)

(21) S = E(a,b) - E(a,b') + E(a',b) + E(a',b')To evaluate equation 20 there is defined Pvv(a,b) in equation 10, but I'm still missing Phh(a,b), Pvh(a,b), and Phv(a,b). How do I get to these three? Is equation 20 supposed to equal cos^2(b-a)?

 

[Nugatory]

The data set for above is for Alice:
a: ++-+--+
b: --+-++-
c: -+-++-+

I have found a violation from data set from above post. Converting above data to horizontal format including Bobs values:

..A....B
a b c...a b c
+ - -...- + +
+ -+...- + -
- + -...+ - +
+ - +...- + -
- + +...+ - -
- + -...+ - +
+ - +...- + -

n[a+c+] + n[b+c-] ≥ n[a+b+]
n=1 + n=1 ≥ n= 4 violation

Written in this form, the inequality applies to the results on one side (or the other), so we have for Alice $3+2\geq{0}$, for Bob $2+3\geq{0}$, and no violation.

That's just a trick of the notation: n[a+c+] means "Alice would measure + in the a and c directions" which means that either Alice measured a+ and Bob c-, or Alice measured c+ and Bob measured a-.

 

[RUTA]

That. So I'm looking at page 907 and I'm trying to calculate their example:

(Ex) a= -45, a'= 0, b= -25.5, b'= 25.5

...they say result is obtained with equations 10, 20, and 21:

(10) Pvv(a,b) = cos^2(b-a) / 2

(20) E(a,b) = Pvv(a,b) + Phh(a,b) - Pvh(a,b) - Phv(a,b)

(21) S = E(a,b) - E(a,b') + E(a',b) + E(a',b')


To evaluate equation 20 there is defined Pvv(a,b) in equation 10, but I'm still missing Phh(a,b), Pvh(a,b), and Phv(a,b). How do I get to these three? Is equation 20 supposed to equal cos^2(b-a)?

First, you caught the missing minus sign for b, good work.

You can get P_HH, P_VH, and P_VH from Eq (7) using <V| and <H| as needed:

P_HH = cos²(α-β)/2
P_HV = P_VH = sin²(α-β)/2.

Of course it must be the case that P_VV + P_HH + P_HV + P_VH = 1.

Now, from Eq (20) we have E(α-β) = cos²(α-β) - sin²(α-β) = cos2(α-β). This gives

E(a,b) = E(-45, -22.5) = cos(-45) = 1/√2
E(a,b') = E(-45, 22.5) = cos(-135) = -1/√2
E(a',b) = E(0, -22.5) = cos(45) = 1/√2
E(a',b') = E(0, 22.5) = cos(-45) = 1/√2.

Thus, Eq (21) gives S = E(a,b) - E(a,b') + E(a',b) + E(a',b') = 2√2 as shown in Eq (23)

 

[Jabbu]

P_HH = cos²(α-β)/2
P_HV = P_VH = sin²(α-β)/2.

Of course it must be the case that P_VV + P_HH + P_HV + P_VH = 1.

So those are counts of matching and mismatching pairs. Equation 20 is then the same what I called correlation formula: match - mismatch, and QM's cos^2(theta). It means two uniformly random binary sequences should be completely uncorrelated with equal number of matches and mismatches.

0.5:0.5 vs 0.5:0.5
chance for match: (0.5 * 0.5) + (0.5 * 0.5) = 0.5
chance for mismatch: (0.5 * 0.5) + (0.5 * 0.5) = 0.5
correlation: 0.5 - 0.5 = 0%

Now, from Eq (20) we have E(α-β) = cos²(α-β) - sin²(α-β) = cos2(α-β). This gives

Instead of theta = (a-b), shouldn't that be (b-a)? So E(a,b) = cos^2(b-a)?

E(a,b) = E(-45, -22.5) = cos(-45) = 1/√2
E(a,b') = E(-45, 22.5) = cos(-135) = -1/√2
E(a',b) = E(0, -22.5) = cos(45) = 1/√2
E(a',b') = E(0, 22.5) = cos(-45) = 1/√2.

How do you get theta = -45 from (-45, -25.5)? Shouldn't that be theta = (-25.5 - (-45)) = 25.5? Also theta = -135 from (-45, 22.5)?


This is what I get:

E(a,b) = E(-45, -25.5) = cos^2(19.5) = 0.89
E(a,b') = E(-45, 25.5) = cos^2(70.5) = 0.11
E(a',b) = E(0, -25.5) = cos^2(-25.5) = 0.82
E(a',b') = E(0, 25.5) = cos^2(25.5) = 0.82

S = 0.89 - 0.11 + 0.82 + 0.82 = 2.42

 

[Jabbu]

I just showed you that a local hidden variables theory can reproduce exactly the predictions of QM for a single filter setting (both Alice and Bob have the filters at the same angle). So what you're saying is false: the predictions for a single angle does not demonstrate that anything nonlocal is going on. In contrast, when you have 3 filter settings to choose from, it is not possible to reproduce the predictions of QM by any local theory.

What you showed are some statements that do not correspond to anything I've seen anywhere else. You used no any equations, you just asserted "we can prove" without showing any proof.

I don't know why you keep bringing up Malus' law. It is completely useless in predicting correlations for the case of entangled photons, because you can't apply it unless you know the photon polarizations.

Because that's how classical physics prediction is calculated. What do you think is classical physics prediction for theta = 0 degrees?

 

[RUTA]

So those are counts of matching and mismatching pairs. Equation 20 is then the same what I called correlation formula: match - mismatch, and QM's cos^2(theta). It means two uniformly random binary sequences should be completely uncorrelated with equal number of matches and mismatches.

0.5:0.5 vs 0.5:0.5
chance for match: (0.5 * 0.5) + (0.5 * 0.5) = 0.5
chance for mismatch: (0.5 * 0.5) + (0.5 * 0.5) = 0.5
correlation: 0.5 - 0.5 = 0%

I don't know what you're talking about, sorry. You have completely correlated outcomes for α-β = 0 (P_VV = P_HH = \frac{1}{2}cos²(0) = \frac{1}{2} and P_VH = P_HV = \frac{1}{2}sin²(0) = 0) and completely anti-correlated outcomes for α-β=\frac{∏}{2} (P_VV = P_HH = \frac{1}{2}cos²(\frac{∏}{2}) = 0 and P_VH = P_HV = \frac{1}{2}sin²(\frac{∏}{2}) = \frac{1}{2}).

Instead of theta = (a-b), shouldn't that be (b-a)? So E(a,b) = cos^2(b-a)?

It doesn't matter, the calculation is the same either way. Your equation for E(a,b) is wrong. E(a,b) = cos²(a-b) - sin²(a-b) = cos2(a-b). That's why your computations of E in the following are wrong.

How do you get theta = -45 from (-45, -25.5)? Shouldn't that be theta = (-25.5 - (-45)) = 25.5? Also theta = -135 from (-45, 22.5)?

This is what I get:

E(a,b) = E(-45, -25.5) = cos^2(19.5) = 0.89
E(a,b') = E(-45, 25.5) = cos^2(70.5) = 0.11
E(a',b) = E(0, -25.5) = cos^2(-25.5) = 0.82
E(a',b') = E(0, 25.5) = cos^2(25.5) = 0.82

S = 0.89 - 0.11 + 0.82 + 0.82 = 2.42

E(a,b) = cos2(a-b) = cos2(-45 + 22.5) = cos(-45) = 1/√2
Etc.

 

[DrChinese]

This is what I get:

E(a,b) = E(-45, -25.5) = cos^2(19.5) = 0.89
E(a,b') = E(-45, 25.5) = cos^2(70.5) = 0.11
E(a',b) = E(0, -25.5) = cos^2(-25.5) = 0.82
E(a',b') = E(0, 25.5) = cos^2(25.5) = 0.82

S = 0.89 - 0.11 + 0.82 + 0.82 = 2.42

You used the wrong angle settings. Instead of 25.5 it should be 22.5. Then the arithmetic works out.

Please stop referring to Malus and classical predictions for entanglement, a non-classical phenomena. This is your last warning on that.

At this point, it is completely unclear what question you are asking. You are switching from context to context so rapidly that none of the responders can assist. So ask a question clearly and immediately that can be addressed, or I suspect this thread will be reaching a conclusion. If you answer anything else instead, please expect the obvious to occur. This is not a forum for debate, or for you to put forth your (misinformed) opinions. We have been quite patient, Jabbu, but that patience is wearing thin. There are multiple professional physicists and science advisors spending time here, and you are going in circles.

 

[stevendaryl]

What you showed are some statements that do not correspond to anything I've seen anywhere else. You used no any equations, you just asserted "we can prove" without showing any proof.

The predictions of QM for a single angle \theta=0 using entangled photons is this:
(Both Alice and Bob keep their filters set at this angle, for each trial).

50\% of the time, one photon passes through Alice's filter and the other passes through Bob's filter.[/itex]
50\% of the time, one photon is blocked by Alice's filter and the other is blocked by Bob's filter.
It never happens that Alice and Bob get different results.

Do you understand that those are the predictions of QM for this case?

Do you really not understand how one could duplicate those predictions without nonlocal interactions?

Suppose that instead of photons, we have slips of paper that messages are written on. Instead of Alice's filter, we have a person who reads one of the slips of paper, and either says "Pass" or "Block". Do you really not see how you could generate slips of paper so that we could guarantee that Alice gets "Pass" 50% of the time, and Bob gets "Pass" 50% of the time, and they always get the same result?

Of course, you can do it. Half of the time, you write "Pass" on both pieces of paper. Half the time you write "Block" on both pieces of paper. That DOES it! It's trivial.

The weird fact is that if instead of one filter setting, Alice and Bob have a choice of three settings, then there is NO way to do it using messages written on pieces of paper.

 

[stevendaryl]

The predictions of QM for a single angle \theta=0 using entangled photons is this:
(Both Alice and Bob keep their filters set at this angle, for each trial).

50\% of the time, one photon passes through Alice's filter and the other passes through Bob's filter.[/itex]
50\% of the time, one photon is blocked by Alice's filter and the other is blocked by Bob's filter.
It never happens that Alice and Bob get different results.

Do you understand that those are the predictions of QM for this case?

Do you really not understand how one could duplicate those predictions without nonlocal interactions?

Suppose that instead of photons, we have slips of paper that messages are written on. Instead of Alice's filter, we have a person who reads one of the slips of paper, and either says "Pass" or "Block". Do you really not see how you could generate slips of paper so that we could guarantee that Alice gets "Pass" 50% of the time, and Bob gets "Pass" 50% of the time, and they always get the same result?

Of course, you can do it. Half of the time, you write "Pass" on both pieces of paper. Half the time you write "Block" on both pieces of paper. That DOES it! It's trivial.

The weird fact is that if instead of one filter setting, Alice and Bob have a choice of three settings, then there is NO way to do it using messages written on pieces of paper.

Jabbu, can you just say whether you agree that the thought experiment in terms of message passing on pieces of paper can be implemented to give the same statistical predictions as QM, in the case of a single filter setting? That conclusion is so trivial, that I really have no idea what you're asking for when you ask for a proof of it.

The harder proof is to show that it CAN'T be done with three filter settings.

 

[stevendaryl]

Because that's how classical physics prediction is calculated. What do you think is classical physics prediction for theta = 0 degrees?

This is not about classical physics. It's about quantum physics. Malus' law has nothing to do with the argument. Absolutely nothing. Your bringing up Malus' law over and over again means that you really, really don't understand what people are saying to you.

 

[Jabbu]

I don't know what you're talking about, sorry. You have completely correlated outcomes for α-β = 0 (P_VV = P_HH = \frac{1}{2}cos²(0) = \frac{1}{2} and P_VH = P_HV = \frac{1}{2}sin²(0) = 0) and completely anti-correlated outcomes for α-β=\frac{∏}{2} (P_VV = P_HH = \frac{1}{2}cos²(\frac{∏}{2}) = 0 and P_VH = P_HV = \frac{1}{2}sin²(\frac{∏}{2}) = \frac{1}{2}).

I'm talking about those terms in equation 20, in an experiment they are not calculate but counted.

It doesn't matter, the calculation is the same either way. Your equation for E(a,b) is wrong. E(a,b) = cos²(a-b) - sin²(a-b) = cos2(a-b). That's why your computations of E in the following are wrong.

If by "cos2(x)" you don't mean to say "cos^2(x)" I think 2 should go inside brackets like this "cos(2x)".

E(a,b) = cos2(a-b) = cos2(-45 + 22.5) = cos(-45) = 1/√2
Etc.

Ok. So how cos^2(theta) fits in? When a= -30 and b= +30 QM predicts correlation = cos^2(60) = 25%, right? So isn't E(-30,30) supposed to be that same "correlation" value?

 

[Jabbu]

You used the wrong angle settings. Instead of 25.5 it should be 22.5.

Thanks.

Please stop referring to Malus and classical predictions for entanglement, a non-classical phenomena.

How else do you compare experimental results with classical physics prediction?

At this point, it is completely unclear what question you are asking.

I'm talking to RUTA. See the paper he posted, or please ask specific questions about whatever it is unclear to you.

 

[Jabbu]

This is not about classical physics. It's about quantum physics. Malus' law has nothing to do with the argument. Absolutely nothing. Your bringing up Malus' law over and over again means that you really, really don't understand what people are saying to you.

To understand practical implications it is paramount to understand how experimental results differ from classical prediction. Please read the paper RUTA posted, or any other paper or article about Bell's theorem. Comparing experimental results with classical prediction is very important part of the analysis.

 

[Jabbu]

Jabbu, can you just say whether you agree that the thought experiment in terms of message passing on pieces of paper can be implemented to give the same statistical predictions as QM, in the case of a single filter setting?

To be more specific instead of hypothetical piece of paper it's better to attribute hidden variables as properties of actual entities, in this case photons and polarizers. Then it's easier to realize real conditions these variables have to satisfy. So to answer your question, unless the paper says "cos^2(theta)" it would be refuted by every other experiment with known relative polarization different than 45 degrees.

But if you insist your hidden variable can fake experimental results for a single theta setting, then you need to realize S = E(a,b) - E(a,b') + E(a',b) + E(a',b') is defined by independent results of four separate theta settings from four separate experiments. Therefore, if your hidden variable can fake each of those experiments individually it will automatically fake the value of S.

http://en.wikipedia.org/wiki/CHSH_inequality

 

[atyy]

To be more specific instead of hypothetical piece of paper it's better to attribute hidden variables as properties of actual entities, in this case photons and polarizers. Then it's easier to realize real conditions these variables have to satisfy. So to answer your question, unless the paper says "cos^2(theta)" it would be refuted by every other experiment with known relative polarization different than 45 degrees.

Bell's theorem can also rule out any modification of the classical laws that are local, so it doesn't just rule out Malus's Law acting on classical photon pairs, but all local alternatives.

Nonetheless, if you would like to try to see if how close Malus's Law and classical photon pairs can come to mimicking the quantum entangled pairs, you can try this. First consider that each side receives 50% vertical and 50% horizontal photons, and that when their polarizers are both vertical, both sides always get the same result. In this case, you can imagine that this result is obtained because the classical source sends out 50% classical pairs with both photons vertically polarized, and 50% classical pairs with both photons horizontally polarized.

Then keeping the same assumption about the classical source, you can apply Malus's Law for other polarizer settings, and carry out the analysis for this classical case to compare with the quantum entangled case.